Rational function extra problems:
8.
Z
3
x
3
+ 18
x
2
+ 33
x
-
2
(
x
2
+ 6
x
+ 10)
2
dx
Begin using integration by parts:
3
x
3
+ 18
x
2
+ 33
x
-
2
(
x
2
+ 6
x
+ 10)
2
=
Ax
+
B
x
2
+ 6
x
+ 10
+
Cx
+
D
(
x
2
+ 6
x
+ 10)
2
3
x
3
+ 18
x
2
+ 33
x
-
2
=
(
Ax
+
B
)(
x
2
+ 6
x
+ 10) +
Cx
+
D
3
x
3
+ 18
x
2
+ 33
x
-
2
=
(
A
)
x
3
+ (6
A
+
B
)
x
2
+ (10
A
+
B
+
C
)
x
+ (10
B
+
D
)
So we have a system of equations:
A=3
6A+B=18
10A+6B+C=33
10B+D=-2
So we have that A=3, B=0, C=3, D=-2
Thus
Z
3
x
3
+ 18
x
2
+ 33
x
-
2
(
x
2
+ 6
x
+ 10)
2
dx
=
Z
3
x
x
2
+ 6
x
+ 10
+
3
x
-
2
(
x
2
+ 6
x
+ 10)
2
dx
Next we can complete the square to see that
x
2
+6
x
+10 = (
x
+3)
2
+1 and use u substitution
where u=x+3,x=u-3, du=dx
So we get 3
Z
u
-
3
u
2
+ 1
du
+
Z
3
u
-
11
(
u
2
+ 1)
2
du
= 3
Z
u
u
2
+ 1
du
-
3
Z
1
u
2
+ 1
du
+ 3
Z
u
(
u
2
+ 1)
2
du
-
11
Z
1
(
u
2
+ 1)
2
du
we can solve several of these with simple substitution
v
=
u
2
+ 1
, dv
= 2
udu
1
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some we need a trig substitution of
u
=
tanθ, du
=
sec
2
θdθ
3
Z
u
u
2
+ 1
du
-
3
Z
1
u
2
+ 1
du
+ 3
Z
u
(
u
2
+ 1)
2
du
-
11
Z
1
(
u
2
+ 1)
2
du
=
3
2
Z
1
v
dv
-
3
Z
1
u
2
+ 1
du
+
3
2
Z
1
v
2
dv
-
11
Z
sec
2
θ
(tan
2
θ
+ 1)
2
dθ
=
3
2
ln
|
v
| -
3 arctan(
u
)
-
3
2
1
v
-
11
Z
sec
2
θ
sec
4
θ
dθ
=
3
2
ln
|
u
2
+ 1
| -
3 arctan(
u
)
-
3
2
1
(
u
2

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- Spring '08
- Bonner
- Calculus, Integration By Parts, 11:11
-
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